By D.R. BATES (Eds.)

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We shall show that φ(χνχ2) must be either symmetric or antisymmetric. The proof depends on the fact that the Hamiltonian H(xltx2) is symmetric; this follows* from the fact that electrons are in distinguishable particles. We first consider an energy E = E0 such that (40) has nondegenerate solutions φ0. This means that if φ0 and φ0' are any two solutions of (H — Ε0)φ = 0 then φ0' = αφ0 where a is a constant. It will be shown later that the ground state of the He atom is an example of such a nondegenerate state.

MiaiM. t)... = 1 and since there are (N — 1)! ) we obtain N , Η Α N Σ χ(*)\ ) = Σ<*\Ηι\«<)· (98) In the sumf ΣΊ/τ^ there are £iV(iV — 1) terms. Using again the fact that D*D is symmetric we obtain Σ ' 1 A)=jN(N-l)[A '12 and, by (95), H^ ^)=2(Z^

Is a permutation with parity p'. By (96) the summand of (97) is zero for (q,s,t,... s'f,... ) and unity for (q,s,t,... ) = (q',s',t'f... ). The number of permutations P is N\ and therefore (A\A) = 1. ) may be permuted to the numbers (1,2,3,. ) by successive interchanges of pairs of numbers. With given (q,s,t,. . ) the number of interchanges is uniquely even or odd; p is even or odd according to whether the number of inter changes is even or odd. 50 M. J. MiaiM. t)... = 1 and since there are (N — 1)!